(a) Find the equation of the circle passing through the origin and whose center lies on the line [tex]x+y=1[/tex].
(b) Find the equation of the circle passing through the points [tex](-6,5)[/tex], [tex](3,-4)[/tex], and [tex](2,1)[/tex].
(c) Find the equation of the circle passing through the points and whose center lies on the line [tex]3x-2y=1[/tex].
Answer (1)
Establish the general equation of a circle ( x − a ) 2 + ( y − b ) 2 = r 2 and use the given condition 3 a − 2 b = 1 .
Formulate equations by substituting the given points ( − 6 , 5 ) , ( 3 , − 4 ) , and ( 2 , 1 ) into the circle equation.
Solve the system of equations to find the center ( a , b ) = ( − 13 15 , − 13 29 ) .
Determine the radius squared r 2 = 169 3445 and express the final equation of the circle: ( x + 13 15 ) 2 + ( y + 13 29 ) 2 = 169 3445 .
Explanation
Set up the problem Let the equation of the circle be ( x − a ) 2 + ( y − b ) 2 = r 2 , where ( a , b ) is the center and r is the radius. We are given that the center lies on the line 3 x − 2 y = 1 , so 3 a − 2 b = 1 . The circle passes through the points ( − 6 , 5 ) , ( 3 , − 4 ) , and ( 2 , 1 ) . Therefore, these points satisfy the equation of the circle.
Formulate equations We have the following equations:
3 a − 2 b = 1
( − 6 − a ) 2 + ( 5 − b ) 2 = r 2
( 3 − a ) 2 + ( − 4 − b ) 2 = r 2
( 2 − a ) 2 + ( 1 − b ) 2 = r 2
Equating equations (2) and (3): ( − 6 − a ) 2 + ( 5 − b ) 2 = ( 3 − a ) 2 + ( − 4 − b ) 2 36 + 12 a + a 2 + 25 − 10 b + b 2 = 9 − 6 a + a 2 + 16 + 8 b + b 2 61 + 12 a − 10 b = 25 − 6 a + 8 b 18 a − 18 b = − 36 a − b = − 2 a = b − 2
Equating equations (3) and (4): ( 3 − a ) 2 + ( − 4 − b ) 2 = ( 2 − a ) 2 + ( 1 − b ) 2 9 − 6 a + a 2 + 16 + 8 b + b 2 = 4 − 4 a + a 2 + 1 − 2 b + b 2 25 − 6 a + 8 b = 5 − 4 a − 2 b − 2 a + 10 b = − 20 − a + 5 b = − 10
Solve for a and b Now we have two equations:
3 a − 2 b = 1
− a + 5 b = − 10
From equation (1), 3 a = 2 b + 1 , so a = 3 2 b + 1 .
Substitute this into equation (2): − 3 2 b + 1 + 5 b = − 10 − 2 b − 1 + 15 b = − 30 13 b = − 29 b = − 13 29
Now, find a :
a = 3 2 ( − 13 29 ) + 1 = 3 − 13 58 + 13 13 = 3 − 13 45 = − 13 15
Solve for r^2 So, a = − 13 15 and b = − 13 29 .
Now, we find r 2 using the point ( 2 , 1 ) :
r 2 = ( 2 − a ) 2 + ( 1 − b ) 2 = ( 2 + 13 15 ) 2 + ( 1 + 13 29 ) 2 r 2 = ( 13 26 + 15 ) 2 + ( 13 13 + 29 ) 2 = ( 13 41 ) 2 + ( 13 42 ) 2 r 2 = 169 1681 + 169 1764 = 169 3445
Write the equation of the circle The equation of the circle is ( x − a ) 2 + ( y − b ) 2 = r 2 , so ( x + 13 15 ) 2 + ( y + 13 29 ) 2 = 169 3445
Final Answer Therefore, the equation of the circle is ( x + 13 15 ) 2 + ( y + 13 29 ) 2 = 169 3445
Examples
Circles are fundamental in many real-world applications, from designing gears and wheels in mechanical engineering to understanding the orbits of planets in astronomy. In architecture, circular designs provide structural stability and aesthetic appeal, while in computer graphics, circles are essential for creating smooth curves and shapes. Understanding the equation of a circle allows engineers, architects, and designers to create and analyze these circular forms effectively.
